What is the solution for the wave equation using u = cos(kx-wt)?

[kiranm]

For a real stretched string, the wave equation is

(partial deriv)^2 (u)/partial deriv t^2 = (T/mu) (partial deriv)^2 (u)/partial deriv x^2 - B/mu(y)

where T is the tension in the string, mu is its mass per unit length and B is its "spring constant".
Show that the wave given by u = cos(kx-wt) is a solution of this equation

I know that v^2 (speed of wave) = T/mu and v^2= w^2/k^2 and 1/v^2= k^2/w^2

What i attempted was:
(partial deriv)^2 (u)/partial deriv t^2= -w^2 cos (kx-wt)
(partial deriv)^2 (u)/partial deriv x^2= -k^2 cos (kx-wt)
so i plugged this into the equation above and ended up with:
By= (cos(kx-wt))*(-Tk^2 + muw^2)
but i don't know how to prove that wave given by u=cos(kx-wt) is a solution of this equation

 

[ideasrule]

For a real stretched string, the wave equation is

(partial deriv)^2 (u)/partial deriv t^2 = (T/mu) (partial deriv)^2 (u)/partial deriv x^2 - B/mu(y)

Are you sure this is right? I don't think the B/mu(y) term is supposed to be there.

 

[kiranm]

Yeah that's what was given in the question but its B/mu * y but it says that B represents the spring constant

 

[kiranm]

and would you know how to derive the dispersion relation w(k), the wave speed v, and the group velocity vg?

 

[kiranm]

for the above where i had cos(kx-wt)= By/(mu*w^2-Tk^2) the denominator mu*w^2 - Tk^2=0 because v^2= w^2/k^2 = T/mu and cross multiply this makes mu*w^2-Tk^2=0

 

[Dan.kl]

im guessing instead of (B/mu)y it should be (B/mu)u

so yeh find partial derivatives of u wrt x and t, sub these in and sub in cos(kx-wt) for u and it all works out

dispersion relation w(k) is √((k^2T + B)/mu)
find v from v = w/k and v_g from v_g = dw/dk